You visited us 3 times! Enjoying our articles? Unlock Full Access!

Basic Buffer Action

The amount of...

Question

The amount of $B a S O_{4}$ formed upon mixing $100$ ml of $20.8 %$ $B a C l_{2}$ solution with $50$ ml of $9.8 %$ $H_{2} S O_{4}$ solution will be:
(Given that molecular weight of $B a = 137, C l = 35.5, S = 32, H = 1$ and $O = 16$ g/mol)

23.3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

11.65

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

30.6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

33.2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $11.65$ g
$100$ ml of $20.8 %$ $B a C l_{2}$ solution $= 20.8$ g $B a C l_{2} =$ $0.1$ mol.
$50$ ml of $9.8 %$ $H_{2} S O_{4}$ solution $= 4.9$ g $H_{2} S O_{4} =$ $0.05$ mol.
The reaction is as follows:
$B a C l_{2} + H_{2} S O_{4} ⇋ B a S O_{4} + 2 H C l$
Since, $H_{2} S O_{4}$ is the limiting reagent, only $0.05$ moles of $B a S O_{4}$ will form.
Hence, $0.05 \times 233 = 11.65$ g of $B a S O_{4}$ is formed.

Suggest Corrections

Similar questions

B a S O_{4}

100

20.8 %

B a C l_{2}

50

9.8 %

H_{2} S O_{4}

= 137

= 35.5

= 32

= 1

= 16

B a C l_{2}

H_{2} S O_{4}

B a S O_{4}

(B a = 137, C l = 35.5, S = 32, H = 1, O = 16)

B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l

N a_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 N a C l

B a C l_{2} (a q)

H_{2} S O_{4}

B a S O_{4}

500

0.25 M N a_{2} S O_{4}

15.0

B a C l_{2}

B a S O_{4}

B a S O_{4}

y \times 10^{- 3}

y

B a = 137, S = 32, O = 16, N a = 23

Join BYJU'S Learning Program